Answer
$$ - 9$$
Work Step by Step
$$\eqalign{
& x = u + 3v,{\text{ }}y = 2u - 3v \cr
& {\text{Find the partial derivatives }}\frac{{\partial x}}{{\partial u}}{\text{ and }}\frac{{\partial x}}{{\partial v}} \cr
& \frac{{\partial x}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {u + 3v} \right] = 1 \cr
& \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {u + 3v} \right] = 3 \cr
& {\text{Find the partial derivatives }}\frac{{\partial y}}{{\partial u}}{\text{ and }}\frac{{\partial y}}{{\partial v}} \cr
& \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {2u - 3v} \right] = 2 \cr
& \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {2u - 3v} \right] = - 3 \cr
& {\text{Calculate the Jacobian}} \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \frac{{\partial x}}{{\partial u}}\frac{{\partial y}}{{\partial v}} - \frac{{\partial y}}{{\partial u}}\frac{{\partial x}}{{\partial v}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting the partial derivatives into }}\left( {\bf{1}} \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left( 1 \right)\left( { - 3} \right) - \left( 2 \right)\left( 3 \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - 3 - 6 \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - 9 \cr} $$
